Minimize the sum calculated by repeatedly removing any two elements and inserting their sum to the Array
Given N elements, you can remove any two elements from the list, note their sum, and add the sum to the list. Repeat these steps while there is more than a single element in the list. The task is to minimize the sum of these chosen sums in the end.
Examples:
Input: arr[] = {1, 4, 7, 10}
Output: 39
Choose 1 and 4, Sum = 5, arr[] = {5, 7, 10}
Choose 5 and 7, Sum = 17, arr[] = {12, 10}
Choose 12 and 10, Sum = 39, arr[] = {22}
Input: arr[] = {1, 3, 7, 5, 6}
Output: 48
Approach: In order to minimize the sum, the elements that get chosen at every step must the minimum elements from the list. In order to do that efficiently, a priority queue can be used. At every step, while there is more than a single element in the list, choose the minimum and the second minimum, remove them from the list add their sum to the list after updating the running sum.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include<bits/stdc++.h>using namespace std;// Function to return the minimized sumint getMinSum(int arr[], int n){ int i, sum = 0; // Priority queue to store the elements of the array // and retrieve the minimum element efficiently priority_queue<int, vector<int>, greater<int> > pq; // Add all the elements // to the priority queue for (i = 0; i < n; i++) pq.push(arr[i]); // While there are more than 1 elements // left in the queue while (pq.size() > 1) { // Remove and get the minimum // element from the queue int min = pq.top(); pq.pop(); // Remove and get the second minimum // element (currently minimum) int secondMin = pq.top(); pq.pop(); // Update the sum sum += (min + secondMin); // Add the sum of the minimum // elements to the queue pq.push(min + secondMin); } // Return the minimized sum return sum;}// Driver codeint main(){ int arr[] = { 1, 3, 7, 5, 6 }; int n = sizeof(arr)/sizeof(arr[0]); cout << (getMinSum(arr, n));}// This code is contributed by mohit |
Java
// Java implementation of the approachimport java.util.PriorityQueue;class GFG{ // Function to return the minimized sum static int getMinSum(int arr[], int n) { int i, sum = 0; // Priority queue to store the elements of the array // and retrieve the minimum element efficiently PriorityQueue<Integer> pq = new PriorityQueue<>(); // Add all the elements // to the prioriry queue for (i = 0; i < n; i++) pq.add(arr[i]); // While there are more than 1 elements // left in the queue while (pq.size() > 1) { // Remove and get the minimum // element from the queue int min = pq.poll(); // Remove and get the second minimum // element (currently minimum) int secondMin = pq.poll(); // Update the sum sum += (min + secondMin); // Add the sum of the minimum // elements to the queue pq.add(min + secondMin); } // Return the minimized sum return sum; } // Driver code public static void main(String[] args) { int arr[] = { 1, 3, 7, 5, 6 }; int n = arr.length; System.out.print(getMinSum(arr, n)); }} |
Python3
# Python3 implementation of the approach# Function to return the minimized sumdef getMinSum(arr, n): sum = 0 # Priority queue to store the elements of the array # and retrieve the minimum element efficiently pq = [] # Add all the elements # to the priority queue for i in range( n ): pq.append(arr[i]) # While there are more than 1 elements # left in the queue while (len(pq) > 1) : pq.sort(reverse=True) # Remove and get the minimum # element from the queue min = pq[-1]; pq.pop(); # Remove and get the second minimum # element (currently minimum) secondMin = pq[-1]; pq.pop(); # Update the sum sum += (min + secondMin); # Add the sum of the minimum # elements to the queue pq.append(min + secondMin) # Return the minimized sum return sum # Driver codeif __name__ == "__main__": arr = [ 1, 3, 7, 5, 6 ] n = len(arr) print(getMinSum(arr, n))# This code is contributed by chitranayal |
C#
// C# implementation of the approachusing System;using System.Collections.Generic;class GFG{ // Function to return the minimized sum static int getMinSum(int[] arr, int n) { int i, sum = 0; // Priority queue to store the elements of the array // and retrieve the minimum element efficiently List<int> pq = new List<int>(); // Add all the elements // to the priority queue for (i = 0; i < n; i++) { pq.Add(arr[i]); } // While there are more than 1 elements // left in the queue while(pq.Count > 1) { pq.Sort(); // Remove and get the minimum // element from the queue int min = pq[0]; pq.RemoveAt(0); // Remove and get the second minimum // element (currently minimum) int secondMin = pq[0]; pq.RemoveAt(0); // Update the sum sum += (min + secondMin); // Add the sum of the minimum // elements to the queue pq.Add(min + secondMin); } // Return the minimized sum return sum; } // Driver code static public void Main () { int[] arr = { 1, 3, 7, 5, 6 }; int n = arr.Length; Console.WriteLine(getMinSum(arr, n)); }}// This code is contributed by avanitrachhadiya2155 |
Javascript
<script>// JavaScript implementation of the approach // Function to return the minimized sum function getMinSum(arr,n) { let i, sum = 0; // Priority queue to store the elements of the array // and retrieve the minimum element efficiently let pq = []; // Add all the elements // to the prioriry queue for (i = 0; i < n; i++) pq.push(arr[i]); // While there are more than 1 elements // left in the queue while (pq.length > 1) { // Remove and get the minimum // element from the queue let min = pq.shift(); // Remove and get the second minimum // element (currently minimum) let secondMin = pq.shift(); // Update the sum sum += (min + secondMin); // Add the sum of the minimum // elements to the queue pq.push(min + secondMin); } // Return the minimized sum return sum; } // Driver code let arr=[1, 3, 7, 5, 6]; let n = arr.length; document.write(getMinSum(arr, n)); // This code is contributed by rag2127</script> |
48
Time Complexity : O(N * log(N))
Auxiliary Space: O(N)
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